\section{Null Space of Poisson Problem}

We simply wanna show when a Poisson problem is given with a homogeneous Neumann boundary condition, there will be a null space which leads to an indefinite system. As been familiar with by everyone, the Poisson problem states as follow

\begin{equation}\nonumber
-\nabla^2 u= f, u \in C^2(\Omega)
\end{equation}

where $\nabla^2$ is the Laplacian operator. We come out the null space by showing the operator $-\nabla^2$ is positive semi-definite. And the nonzero subspace $\hat{\textbf{U}}$ satisfying $L^2<-\nabla^2\hat{\textbf{U}},\hat{\textbf{U}}> = 0$ spans the null space.

\begin{equation}\nonumber
\begin{split}
L^2<-\nabla^2 u, u> &= \int_{\Omega}<-\nabla^2 u, u>d\Omega = -\oint_{\partial\Omega}u<\nabla u, \vec{\textbf{n}}>ds + \int_{\Omega}<\nabla u, \nabla u>d\Omega\\
&= -\oint_{\partial\Omega}u\frac{\partial u}{\partial \vec{\textbf{n}}}ds + \int_{\Omega}<\nabla u, \nabla u>d\Omega
\end{split}
\end{equation}

Here notice if we apply homogeneous Neumann boundary condtion, $\frac{\partial u}{\partial \vec{\textbf{n}}} = 0$ on $\partial \Omega$, the above equation becomes $\int_{\Omega}<\nabla u, \nabla u>d\Omega$ which is always equal or larger than zero. The zero case happens when $\nabla u = \textbf{0}$ on $\Omega$, which means u is constant over the whole domain. As we just enforce Neumann boundary condition on $\partial \Omega$, u could be any constant function.  Thus we have

\begin{equation}
ker\{-\nabla^2\} = C, C \in \mathbb{R}
\end{equation}